At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.400 M.

Question

N2(g)+O2(g)↽−−⇀2NO(g)

If more NO is added, bringing its concentration to 0.700 M, what will the final concentration of NO be after equilibrium is re‑established?

Answer 1

We can first calculate the equilibrium constant of this reaction since we are given the concentrations of the reactants and the product.

K_c = [NO]²/([N₂][O₂]) = (0.4)²/(0.1*0.1) = 16

However, since more NO is being added to the vessel, we know that Le Chatelier's principle asserts that the reaction will proceed backwards towards the side of the reactants to re-establish chemical equilibrium. Therefore we know that the concentration of the product (NO) will decrease by x, whereas the concentrations of the reactants (N₂ and O₂) will increase by 0.5x proportionally to the decrease of NO. Thus, the equation can be rewritten as follows:

(0.7-x)²/(0.1+0.5x)² = K_c = 16

We can then solve for x using the quadratic formula, from which we calculate x = 0.1.

We can then plug this value of x = 0.1 into the numerator in the equation above to find the concentration of NO:

[NO] = 0.7-0.1 =0.6M